Re: declare -p doesn't show the global attribute

[isabella parakiss]

On 4/27/15, Chet Ramey <chet.ramey@case.edu> wrote:
> On 4/26/15 5:26 PM, isabella parakiss wrote:
>> $ fn () { declare -g var=x; declare -p var; } ; fn
>> declare -- var="x"
>>
>> I think the correct output should be declare -g var="x"
>> Is this intended or is it a bug?
>
> There is no such thing as `the global attribute'.  The -g option simply
> causes declare to create variables at the global scope instead of in a
> function-local scope.  The output you see is no different than what
> would have been displayed had `var' been declared and given a value at
> the global scope outside the function.
>

Ok I understand, but would it be possible to add it?

My use case may not be the best possible example, but I am sourcing a
"library" to set various parts of my interactive shells.  In this library
there's a function that sets the color variables from the output of tput.
Trying to avoid to call tput when it's not needed, I decided to save them
to an external file and source it if it exists.  My problem is that if I
source that file from the function, I'm setting local variables.  This is
the part that sets those variables:
https://gist.githubusercontent.com/izabera/d8f0f63fa3dc168af7b9/raw/581753078d33acea3d28b005722edb42c717fa96/gistfile1.sh

It seems to me that if declare -p showed the -g, I could just set them as
global and readonly before I save them, and I wouldn't need to set them
after I source them, in line 6.

Instead of declare -p, one could use something like printf %s=%q but it's
kinda hard to save arrays that way...


It's not a big deal, I just think it would make more sense.


---
xoxo iza